[Music] hey everyone welcome to PL Academy hope you found this video useful if you did it would be awesome if you could subscribe like the video share it with your friends and maybe drop a comment your support really helps me make more videos in this video I covered everything on the syllabus like you can see in the figure principle of superposition superposition is the principle that when waves of the same type meet at a point they individual displacements combined to produce a resultant displacement as a blue wave travels to the left and a red wave travels to the right initially they are in Phase meaning their phase difference is zero when these waves meet their individual displacements combine constructively resulting in a maximum resultant displacement that is twice the amplitude of either individual wave this is called constructive superposition which occurs when waves are in Phase with phase differences of 0 Dee creating the maximum amplitude when a blue wave travels One Division to the left and a red wave travels One Division to the right they individual displacements combined to create the resultant displacement as shown when a blue wave travels One Division to the left and a red wave travels One Division to the right the phase difference between two waves is pi radians or 180° they are said to be in antiphase their individual displacements combined destructively resulting in a minimum resultant displacement which can be zero if the waves have equal amplitudes this is called destructive superposition which occurs when waves are in anti phase with phase differences of Pi radians or 180° creating the zero amplitude when a blue wave travels One Division to the left and a red wave travels One Division to the right their individual displacements combined to create the resultant displacement as shown when a blue wave travels One Division to the left and a red wave travels One Division to the right the phase difference between two waves is 2 pi radians or 360° they are said to be in Phase their individual displacements combine constructively resulting in a maximum resultant displacement that is twice the amplitude of either individual wave this is called constructive superposition which occurs when waves are in Phase with phase differences of 2 pi radians or 360° creating the maximum amplitude therefore we can be concluded that the constructive superposition which occurs when waves are in Phase with phase differences of 0 2 pi 4 Pi 6 pi and so on the destructive superposition which occurs when waves are antiphase with phase differences of Pi 3 Pi 5 Pi 7 pi and so on exam style question one the three waves shown in each diagram have the same amplitude and frequency but different phase they are added together to give a resultant wave in which case is the resultant wave zero at this instant in a choice a there are three waves as blue violet and green coolers like this at this point displacements of three waves combined to be zero like this at this point displacements of three waves combined to be zero like this at at this point displacements of three waves combined to be zero like this and any positions the resultant displacements of three waves combined together to become zero like this so a is correct stationary wave stationary waves or standing waves are formed by the superposition or interference of two Progressive waves traveling in opposite directions with the the same frequency wavelength and speed this interference results in a pattern of nodes and antinodes at nodes are the destructive superposition their phase differences are Pi 3 Pi 5 Pi 7 pi and so on or 180° 540° 900° 1, 1260° and so on their path differences are Lambda over 2 3 Lambda over 2 5 Lambda over 2 7 Lambda over 2 and so on at anodes are the constructive superposition their phase differences are 0 2 pi 4 Pi 6 pi and so on or 0° 360° 720° 1, 80° and so on their path differences are 0 Lambda 2 lamb LDA 3 Lambda and so on when a blue wave travels to the right and a red wave travels to the left as shown they are moving in opposite directions and they have same frequency wavelength and speed at the initial they are in phase and phase difference is zero they superpose or interfere with each other to create the stationary wave shown in Black Wave After 1/8 of a period each wave travels 1/8 of a wave length in opposite directions resulting in a separation of 1/4 of a wavelength they superpose to form the stationary wave shown in Black Wave After 1/4 of a period each wave travels 1/4 of a wavelength in opposite directions resulting in a separation of half the wavelength they superpose to form the stationary wave shown in Black Wave After 38 of a period each wave travels 38 of a wavelength in opposite direction CS resulting in a 34s wavelength in separation they superpose to form the stationary wave shown in Black Wave after half of period each wave travels half a wavelength in opposite directions resulting in a full wavelength separation they superpose to form the stationary wave shown in Black Wave when two waves traveling at high speeds in opposite directions superpose they can form stationary waves which appear as a series of Loops as shown this creates a pattern of nodes and antinodes as shown at nodes the amplitude is at minimum in this diagram it is zero because the waves have equal amplitudes at antinodes the amplitude is maximum the distance between two Loops is equal to wavelength Lambda so the distance between consecutive nodes or antinodes are half of the wavelength exam style question two the diagram shows a sketch of a wave pattern over a short period of time which description of this wave is correct in the diagram shown the stationary wave the distance for two Loops is a wavelength for one meter has five Loops which is equivalent to 2.5 wavelengths so we can determine the wavelength of the wave is equal to 1 m / 2.5 which is 0.4 M or 40 cm the transverse waves and longitudinal waves can be occurred the stationary wave in the choice a the wave is longitudinal has a wavelength of 20 cm and is stationary this is incorrect in the choice B the wave is transverse has a wavelength of 20 cm and is stationary this is incorrect in the choice C the wave is transverse has a wavelength of 40 cm and is Progressive this is incorrect in the choice d The Wave is transverse has a wavelength of 40 cm and is stationary this is correct stationary wave of the stretched string a vibrator generates an incident wave that travels to the left this wave reflects at the pulley creating a reflected wave traveling to the right the incident and reflected waves have the same frequency wavelength and speed these two waves superpose to produce a standing wave the simplest standing wave pattern with a single Loop is called the fundamental frequency or the first harmonic the fixed ends at the vibrator and pulley are always formed nodes because the string cannot free to vibrate the vibrator varies the frequency until a stationary wave with two Loops second harmonic three Loops third harmonic four Loops fourth harmonic and so on let L be the length of the stretched string recall that two Loops of stationary wave equals one wavelength for first harmonic has a single Loop indicating that L equals half of wavelength so the wavelength equals 2 L the fundamental frequency F1 can be calculated by dividing the speed V by wavelength Lambda substituting Lambda equal 2 L so fub1 = V / 2L 4 second harmonic has two Loops indicating that L equals wavelength Lambda the second harmonic frequency FS2 is calculated by dividing the speed V by wavelength Lambda substituting Lambda equals l so FS2 = V / l or 2 v/ 2 L and V / 2 L = fub1 so FS2 is twice as fub1 43d harmonic has three Loops indicating that L equals 3 Lambda / 2 so Lambda equal 2 L over 3 the third harmonic frequency F3 is calculated by dividing the speed V by wavelength Lambda substituting Lambda = 2 L over 3 so F3 = 3 V / 2 L and V / 2 L equals FS1 so F3 is three times of fub1 for fourth harmonic has four Loops indicating that L equal 2 Lambda so Lambda equal L / 2 the fourth harmonic frequency F3 is calculated by dividing the speed V by wavelength Lambda substituting Lambda = L / 2 so F4 = 2 V / l or 4 V / 2 L and V / 2 L = fub1 so F4 is fourth times of fub1 exam style question three a stationary wave is produced on a string that is stretched between two fixed points that are a distance of 1.35 M apart as shown the speed of the waves on the string is 450 m/s what is the frequency of oscillation of the stationary wave the standing wave has three Loops which corresponds to 1.5 wavelengths so 1.35 m equal 1.5 wavelengths this means one wavelength is 1.35 m / 1.5 is equal to 0.9 M the frequency of the stationary wave can be calculated by the equation FAL v/ Lambda substituting V equal 450 m/s and Lambda equal 0.9 M we get the frequency equals 500 htz so C is correct the stationary wave of sound wave in air column with close one end a speaker generates an incident sound wave that travels through a pipe closed at one end this wave reflects at the closed end creating a reflected wave traveling in the opposite direction the incident and reflected waves have the same frequency wavelength and speed the superposition of these two waves can produce a standing wave the simplest standing wave with a quarter Loop pattern is called the fundamental wave or the first harmonic this corresponds to the lowest frequency that can be heard hence the term fundamental frequency a closed end of the pipe forms a node because air particles cannot vibrate freely an open end forms an antinode because air particles can vibrate freely the speaker varies the frequency until a stationary wave with second her second harmonic three herd third harmonic and so on let L be the length of the air column recall that two Loops of stationary wave equals one wavelength for first harmonic has an one4 Loop pattern indicating that L equals 1/4 of a wavelength so the wavelength equals 4 L the fundamental frequency fub1 can be calculated by dividing the speed V by wavelength Lambda substituting Lambda equal 4 L so fub1 = v/ 4 L 4C harmonic has a 3/4 Loop pattern indicating that L equals 3/4 of a wavelength so the wavelength equals 4 L over 3 the second harmonic frequency FS2 can be calculated by dividing the speed V by wavelength Lambda substituting Lambda equals 4 L over 3 so F2 = 3 V over 4 L and V over 4 L = fub1 so FS2 = 3 * of fub1 43 harmonic has a 54 Loop pattern indicating that L equals 5/4 of a wavelength so the wavelength equals 4 L the third harmonic frequency F3 can be calculated by dividing the speed V by wavelength Lambda substituting Lambda equals 4 L over 5 so F3 = 5 V over 4 L and V over 4 L = fub1 so F3 = 5 * of fub1 exam style question four a long glass tube is almost completely immersed in a large tank of water a tuning fork is struck and held just above the open end of the tube as it is slowly raised a louder sound is first heard when the height h of the end of the tube above the water is 18.8 CM a louder sound is next heard when H is 56.4 cim the speed of sound in air is 330 m/s what is the frequency of the sound produced by The Tuning Fork the first herd when H is 18.8 CM the next herd when H is 56.4 CM so half of wave length equal 56.4 - 18.8 is equal to 37.6 CM therefore 1 wavelength equal 2 * 37.6 is equal to 0.752 M the frequency can be calculated using the equation FAL V / Lambda substituting V = 330 m/s and Lambda equal 0.7 5 2 m we get the frequency F equal 440 Hertz for two significant figures so B is correct the stationary wave of sound wave in air column with open both ends a speaker generates an incident sound wave that travels through an open pipe this wave reflects at the open end creating a reflected wave traveling in the opposite direction the incident and reflected waves have the same frequency wavelength and speed the superposition of these two waves can produce a standing wave the simplest standing wave with one Loop pattern is called the fundamental wave or the first harmonic this corresponds to the lowest frequency that can be heard hence the term fundamental frequency both open ends forms an antinode because air particles can vibrate freely the speaker varies the frequency until a stationary wave with second her second harmonic three herd third harmonic and so on let L be the length of the air column recall that two Loops of stationary wave equals one wavelength for first harmonic has an one Loop pattern indicating that L equals half of a wavelength so the wavelength equals 2 L the fundamental frequency fub1 can be calculated by dividing the speed V by wavelength Lambda substituting Lambda equals 2 L so fub1 = V / 2 L 4 second harmonic has a two Loops pattern indicating that L equals one wavelength the second harmonic frequency F2 can be calculated by dividing the speed V by wavelength Lambda substituting Lambda equals l so F2 = V / l or 2 V / 2 L and v/ 2 L = fub1 so FS2 = 2 * of fub1 43 harmonic has a three Hales Loops pattern indicating that L equals 3 halves of a wavelength so the wavelength equals 2 L over 3 the third harmonic frequency F3 can be calculated by dividing the speed V by wavelength Lambda substitute Lambda = 2 l/ 3 so F3 = 3 V / 2 L and v/ 2 L = fub1 so F3 = 3 * of fub1 exam style question five stationary sound waves can be formed in the air Columns of pipes one type of pipe is closed at one end and open at the other end another of pipe is open at both ends which pipe can form a stationary sound wave with the lowest frequency let V represent the speed of sound in air Choice a consider a pipe open at both ends the lowest frequency or fundamental frequency standing wave has a half wavelength pattern one Loop is shown one Loop is equivalent to half wavelength so 2 L equals a half of Lambda and Lambda equal 4 L using the equation the frequency F equal speed V over wavelength Lambda substituting Lambda equals 4 L so F = 0.25 V over L Choice B consider a pipe is closed at one end the lowest frequency or fundamental frequency standing wave has a quarter wavelength pattern half Loop is shown a half Loop is equivalent a quarter wavelength so 2 L = 1/4 Lambda and Lambda equal 8 L using the equation the frequency F equal speed V over wavelength Lambda substituting Lambda equal 8 l so F = 0.125 V / L Choice C consider a pipe open at both ends the lowest frequency or thund mental frequency standing wave has a half wavelength pattern one Loop is shown one Loop is equivalent to half wavelength so L equals a half of Lambda and Lambda equal 2 L using the equation the frequency FAL speed V over wavelength Lambda substituting Lambda equal 2 L so F equal 0.5 V / L Choice d consider a pipe is closed at one end the lowest frequency or fundamental frequency standing wave has a quarter wavelength pattern half Loop is shown a half Loop is equivalent to quarter wavelength so L equal 1/4 Lambda and Lambda equal 4 L using the equation the frequency F equal speed V over wavelength Lambda substituting Lambda equals 4 L so F equal 0.25 V / l so B has the lowest frequency stationary wave of microwave an incident microwave traveling toward a metal reflector is reflected and travels in the opposite direction the incident and reflected waves have the same frequency wavelength and speed the superposition of the incident and reflected microwaves creates a stationary wave pattern with nodes and antinodes a detector measures the microwave signal intensity of microwave this detector can be used to locate the nodes and anten noes of this standing wave by moving the detector between the microwave source and the reflector the alternating points of minimum and maximum signal intensity can be observed the points of minimum signal intensity correspond to the nodes the points of Maximum signal intensity correspond to the ant noes the distance between successive nodes or antinodes is equal to half of a wavelength exam style question six in an experiment to demonstrate a stationary wave two microwave transmitters emitting waves of wavelength for cmers are set facing each other as shown a detector is moved along a straight line between the transmitters it detects positions of Maximum and minimum signal the detector is a distance D from the leftand transmitter assume that both transmitters are at antinodes of the stationary wave which row gives a value of D for a maximum and for a minimum at the maximum signals is the antinodes at the minimum signals is the nodes the the question given the wavelength of microwave is 4 cm so half of a wavelength is 2 cm the distance between consecutive antinodes or nodes is 2 cm and the distance between consecutive antinodes and nodes is 1 cenm from the diagram shows that the value of D for a maximum signals or antinodes occur to be 0 2 4 6 8 and so on which are even numbers the value of D for minimum signals or nodes occur to be 1 3 5 7 and so on which are odd numbers so the value of D for a maximum can be 46 and 48 CM the value of D for a minimum can be 47 CM so C is correct difference between the stationary wave and the progressive wave the progressive wave can be transferred the energy from one place to another the progressive wave transfers the energy in this direction the stationary wave cannot be transferred the energy from one place to another however the energy is stored in the vibrating particles like this all the points along the progressive wave have the same amplitude these amplitude is the maximum displacement that all point in wave can be oscillated the stationary wave has the amplitude of different point to vary from a maximum to zero the zero amplitude is at nodes the maximum amplitude is at antinodes all the points over one wavelength in Progressive wave have different phases all the points between successive nodes are in phase for example all points in same color section her phase difference is zero all points in pink and yellow region have phase difference equals 360° which is also in Phase all points in red and yellow regions have phase difference equals 720° which is also in Phase all points in blue and yellow regions have phase difference equals 180° which is antiphase all points in green and yellow regions have phase difference equals 540° which is antiphase exam style question seven which statement concerning a stationary wave is correct Choice a all the particles between two adjacent nodes oscillate in Phase this is correct Choice B the amplitude of the stationary wave is equal to the amplitude of one of the Waves creating it this is incorrect the amplitude of a standing wave varies from0 Z at the nodes to a maximum which can be twice the amplitude of the individual waves at the anten noes Choice C the wavelength of the stationary wave is equal to the separation of two adjacent nodes this is incorrect the distance between two adjacent nodes is half a wavelength Choice D there is no displacement of a particle at an antinode at any time this is incorrect at antinodes it's displacement is maximum while no displacement at [Music] nodes hey everyone welcome to PL Academy hope you found this video useful if you did it would be awesome if you could subscribe like the video share it with your friends and maybe drop a comment your support really helps me make more videos in this video I covered everything on the syllabus like you can see in the figure defraction of wave defraction is the spreading out of a wave as it passes through a narrow Gap aperture or around an obstacle the amount of defraction depends on the wavelength of the wave compared with the size of the Gap aperture or the size of the obstacle when straight wave fronts in encounter a gap or aperture in a barrier they pass through and defract spreading out into the region Beyond The Wave fronts curve at the edges of the Gap while the central portion remains relatively straight this shown that these area are diffracted this area is not defract as the size Gap decreases to approach the wavelength defraction becomes more resulting in Greater spreading and curvature of the wave fronts as shown as the size Gap decreases to very smaller than the wavelength defraction increases resulting in Greater spreading and curvature of the wave fronts but the defraction pattern may be become less distinct as shown so we can conclude as follows when the aperture is much larger than the wavelength defraction is minimal Little Wave forms curvature when the aperture size is comparable to the wavelength defraction increases greater curvature of wave fronts when the aperture is much smaller than the wavelength defraction is significant but the defraction pattern may become less distinct when a wave encounters an obstacle defraction occurs around the edges when straight wave fronts with a wavelength much smaller than the width of the obstacle defraction is minimal resulting in slight curvature of the wave fronts at the edges as shown this creates a shadow region behind the obstacle where the wave is blocked as wavelength increases defraction increases leading to Greater curvature of the wave fronts at the edges as shown this results in a less distinct or no Shadow region behind the obstacle as the wave bends around it when a sound wave passes through a doorway the size of the doorway is comparable to the wavelength of the sound causing the defraction to occur when light passes through a doorway the size of the doorway is typically much larger than the wavelength of light resulting in no defraction to investigate the defraction of the visible light using a single slit the slit width should be less than approximately 10 power -4 M or 0.1 mm this is because the wavelength of visible light ranges from about 10^ -6 to 10 power -7 M when monochromatic light from a laser Source passes through a single slit it diffracts as shown the central Fringe of the defraction pattern has the highest intensity and greatest width decreasing the slit width increases defraction leading to Greater spreading of the light and a decrease in intensity this increased defraction also results in a wider Central Fringe the defraction pattern of white light through a single slit is shown the central Fringe is white while the other fringes display a spectrum of colors with blue light closer to the center and red light further away this occurs because red light has a longer wavelength than blue light and longer wavelengths defract more spread out more the defraction patterns of red green and blue light through a single slit are also shown the red lights Central Fringe is the widest while the blue lights Central Fringe is the narrowest this is consistent with red lights longer wavelength which causes greater defraction than green or blue light to investigate the defraction of water waves using a single slit in a ripple tank the wave fronts of waves are generated using a vibrating bar and move towards a gap in a barrier where the Ripple stri strike the barrier they are reflected back where they arrive at the Gap however they pass through and spread out into the space beyond it is this spreading out of waves as they travel through a gap or past the edge of a barrier a ripple tank is a shallow tray of water with a light source shining down through it the light illuminates the wave fronts making them visible a straight Dipper can be used to create straight wave fronts in a ripple tank when the Dipper is vibrated up and down it creates a series of parallel wave fronts passing through a gap to observe the defraction of the water wave on a screen the stroboscope or video camera is used to make the slow motion of the wave fronts of water wave on the screen exam style question one the speed of sound in air is 330 m/s which size of architectural features in a large Concert Hall would best defract sound waves of frequency 0.44 khz the size of architecture features should be comparable to wavelength of the sound waves resulting in the best defraction so we can calculate the wavelength of the sound first using the equation wavelength Lambda equals speed V over frequency F substituting V equal 330 m/ second and f = 0.44 * 10^ 3 Hertz we get the wavelength Lambda equal 0.75 m or 750 mm so the size of architecture features should be 750 mm exam style question two in an experiment water waves in a ripple tank are incident on a gap as shown some defraction of the water waves is observed which change to the experiment would provide a better demonstration of defraction the best defraction occurs when the size of Gap is comparable to the wavelength Choice a increase the amplitude of the Waves this is incorrect because the amplitude does not affect defraction Choice B increase the frequency of the Waves this is incorrect because increasing the frequency decreases the wavelength which would reduce defraction Choice C increase the wavelength of the Waves this is correct because increasing the wavelength makes it closer to the gap size leading to more defraction Choice D increase the width of the Gap this is incorrect because increasing the Gap width decreases the amount of defraction [Music] hey everyone welcome to PL Academy hope you found this video useful if you did it would be awesome if you could subscribe like the video share it with your friends and maybe drop a comment your support really helps me make more videos in this video I covered everything on the syllabus like you can see in the figure interference of wave inter interference occurs when two coherent waves meet at a point they superpose producing a pattern of nodes and antinodes at the antinodes have the phase differences to be 0er 2 pi 4 Pi 6 pi and so on or 0 360° 720° and so on and their patch differences are zero Lambda 2 Lambda 3 Lambda and so on at the nodes have the phase differences to be Pi 3 Pi 5 Pi 7 pi and so on or 180° 540° 900° and so on and their patch differences are half a Lambda 3 half Lambda 5 half Lambda and so on waves are considered coherent when they satisfy the following conditions they have a constant phase difference this is the most important condition they are the same type of wave they have the same frequency and wavelength for example green and blue waves are coherent this is because both waves maintain a constant phase difference of 180° and they have the same frequency and wavelength their amplitude are not equal but do not affect coherence pink and blue waves are coherent this is because both waves maintain a constant phase difference of 90° and they have the same frequency and wavelength their amplitude are not equal but do not affect coherence red and green waves are not coherent this is because they do not maintain a constant phase difference and do not have the same frequency or wavelength to investigate the interference of water waves in the Ripple tank two Dippers connected to the same oscillator vibrate with the same frequency generating two coherent water waves when the bar vibrates each Dipper acts as a source of circular ripples spreading outwards where these sets of ripples overlap an interference pattern is projected onto the screen below this interference pattern is visible on the screen the pattern consists of alternating regions of constructive and destructive interference along this line are locations of destructive interference resulting in nodes along this line are locations of constructive interference resulting in antinodes along this line are locations of destructive interference resulting in nodes along this line are locations of constructive interference resulting in antinodes along this line are locations of destructive interference resulting in nodes along this line are locations of constructive interference resulting in antinodes along this line are locations of destructive interference resulting in nodes another way to observe interference in a ripple tank is to use plain waves passing through two gaps in a barrier the water waves are defract at the gaps and then interfere Beyond creating a similar pattern the interference pattern in the Ripple tank two sources S1 and S2 produce coherent waves with a phase difference of zero the waves have the same wavelength Lambda given the red line represents the crest and green line represents the trough along the yellow lines are locations of constructive interference resulting in antinodes these points are formed the superposition of wave crests with crests or wave trough with troughs along this black lines are locations of destructive interference resulting in nodes these points are formed the superposition of wave crests with troughs recall that at the nodes have the phase differences to be Pi 3 Pi 5 Pi 7 pi and so on or 180° 540° 900° and so on and their patch differences are half Lambda 3 half Lambda 5 half Lambda and so on at the antinodes have the phase differences to be zero 2 pi 4 Pi 6 pi and so on or zero 360° 720° and so on and their patch differences are zero Lambda 2 Lambda 3 Lambda and so on the difference in distance from any point along black and yellow lines to sources S1 and S2 respectively is the path difference for that line the central yellow line marks the zero order antinode a0 at this antinode both the path difference and the phase difference are zero this means that the distance from any point on this line to the two sources S1 and S2 is equal for example at point a on this line the distance S1 to a is 5 Lambda and the distance S2 to a is also 5 Lambda at the left next black line marks the first order node N1 at this node the path difference is half a wavelength and the phase difference are 180° this means that the difference in distance from any point on this line to the two sources S1 and S2 is half a wavelength for example at point B on this line the distance S1 to B is 5 Lambda and the distance S2 to B is also 5.5 Lambda at the next yellow line marks the first order anode A1 at this anode the path difference is one wavelength and the phase difference are 360° this means that the difference in distance from any point on this line to the two sources S1 and S2 is one wavelength for example at Point C on this line the distance S1 to C is 4.5 Lambda and the distance S2 to C is also 5.5 Lambda at the next black line marks the second order node N2 at this node the path difference is three half wavelength and the phase difference are 540° this means that the difference in distance from any point on this line to the two sources S1 and S2 is three half wavelength for example at Point D on this line the distance S1 to D is 3 Lambda and the distance S 2 to D is also 4.5 Lambda at the next yellow line marks the second order antinode A2 at this antinode the path difference is two wavelength and the phase difference are 720° this means that the difference in distance from any point on this line to the two sources S1 and S2 is two wavelength for example at Point e on this line the distance S1 to e is 4 Lambda and the distance S 2 to e is also 6 Lambda at the next black line marks the third order node N3 at this node the path difference is five half wavelength and the phase difference are 900° this means that the difference in distance from any point on this line to the two sources S1 and S2 is five half wve length for example at Point F on this line the distance S1 to F is 3 l Lambda and the distance S2 to F is also 5.5 Lambda at the next yellow line marks the third order antinode A3 at this anode the path difference is three wavelength and the phase difference are 1, 180° this means that the difference in distance from any point on this line to the two sources S1 and S2 is three wavelength for example at Point G on this line the distance S1 1 to G is 2.5 Lambda and the distance S2 to G is also 5.5 Lambda this pattern is symmetric about the central yellow line like this exam style question one two sources of microwaves p and Q produce coherent waves with a phase difference of 180° the waves have the same wavelength Lambda and at the point s there is a minimum in the interference pattern produced by waves from the two sources the distance Qs minus PS is called the path difference which expression could represent the path difference at the initial the waves from sources p and Q have phase difference of 180° so the path difference between two waves is half a wavelength the path difference for minimum interferences at Point s can be one half Lambda 3 halfes Lambda 5 Hales Lambda and so on the half Lambda path difference from the sources adds to any path difference to determine the overall path difference at Point s Choice a the overall path difference = 1/4 Lambda + 12 Lambda = 3/4 Lambda this is incorrect Choice B the overall path difference equal 1 12 Lambda + 1 12 Lambda = 1 lamb Lambda this is incorrect Choice C the overall path difference equals 1 Lambda + 1 12 Lambda equal 3es Lambda this is correct Choice D the overall path difference equals 3es Lambda + 1 12 Lambda = 2 Lambda this is incorrect exam style question two an outdoor concert has two large speakers beside the stage for broadcasting music in order to test the speakers they are made to emit sound of the same wavelength and the same amplitude the curved lines in the diagram represent wave fronts where is the loudest sound heard the wave fronts connect points of equal phase in the sound wave such as compressions the loudest sound occurs where constructive interference takes place constructive interference happens when compressions from both speakers overlap or when rare factions from both speakers overlap this occurs where the wave fronts from the two speakers intersect so D is correct investigating the interference of sound waves two identical loudspeakers connected to a single generator produce coherent sound waves with a phase difference of zero as a microphone is moved along line AB in front of the loudspeakers a series of alternating loud and quiet regions is observed at the loud regions constructive interference occurs resulting in a sound intensity that is twice that of a single loudspeaker these regions are called antinodes at the quiet regions destructive interference occurs resulting in a sound intensity that is lower than that of a single loudspeaker these regions are called nodes the the central loud region marks the zero order antinode at this antinode both the path difference and the phase difference are zero at the next quiet region marks the first order node at this node the path difference is half a wavelength and the phase difference are Pi radians at the next loud region marks the first order antinode at this antinode the path difference is one wavelength and the phase difference are 2 pi radi Ian at the next quiet region marks the second order node at this node the path difference is three half wavelength and the phase difference are 3 Pi radians this pattern is symmetric about the central quiet like this the graph shows the variation of sound intensity detected by the microphone with distance from point A as displayed on the C the intensity at nodes may not be exactly zero due to the presence of background noise exam style question three two loudspeakers are placed near to each other and facing in the same direction a microphone connected to an oscilloscope is moved along a line some distance away from the loudspeakers as shown which statement about the waves emitted by the loudspeakers is not a necessary condition for the microphone to detect a fixed point along the line where there is no sound Choice a the waves must be emitted in Phase this is not a necessary because the interference can be occurred if the waves emitted in out of phase Choice B the waves must be emitted with a similar amplitude this is a necessary because we need to detect where no sound which means that completely destructive interference Choice C the waves must have the same frequency this is necessary because it is condition of coherent waves Choice D the waves must have the same wavelength this is necessary because it is condition of coherent waves investigating the interference of light waves a white light source is used rather than a laser a monochromatic filter allows only one wavelength of light to pass through a single slit diffracts the light this defract light arrives in phase at the double slit which ensures that the two parts of the double slit behave as coherent sources of light the double slit is placed a centimeter or two beyond the single slit acts as two coherent sources the interference pattern is observed on a screen about a meter away the bright fringes have equal intensity and are equally fringes separation as shown the experiment has to be carried out in a darkened room as the intensity of the light is low and the fringes are hard to see when a red filter is used a interference pattern is shown when a green filter is used a interference pattern is shown showing that the Fringe separation is smaller than with red light when a blue filter is used a interference pattern is shown showing that the Fringe separation is smaller than with green light when no filter the interference pattern of white light is shown this pattern shows a spectrum cooler with blue light fringes closer to the center and red light fringes further away this is because red light has a longer wavelength than blue light and longer wavelengths defract more spread out more three key factors influence the experimental setup slit width slit widths are typically a fraction of a millimeter since visible light wavelengths are much smaller minimal defraction occurs allowing for sufficient light intensity to observe fringes narrower slits would significantly reduce light intensity slit separation double slits are usually separated by about a millimeter larger separations would result in closely spaced fringes making them difficult to distinguish screen distance the screen is placed approximately 1 M from the slits this distance provides clearly separated fringes without excessive dimming of the light exam style question four light of wavelength Lambda is incident normally on two narrow slits S1 and S2 a small distance apart bright and dark fringes are observed on a screen a long distance away from the slits the N dark Fringe from the central bright Fringe is observed at Point p on the screen which equation is correct for all positive values of n all positive values of n equal 1 2 3 4 and so on the dark fringes occur where destructive interference take place which happens when the path differences are 1 half Lambda 3 Hal Lambda 5 Hales Lambda and so on Choice a the path differences are 1 half Lambda Lambda 3es Lambda and and so on this is incorrect Choice B the path differences are Lambda 2 Lambda 3 Lambda and so on this is incorrect Choice C the path differences are 1 half Lambda 3 half Lambda 5 Hales Lambda and so on this is correct Choice D the path differences are three halfes Lambda 5 Hales Lambda 7 Hales Lambda and so on this is incorrect Young's double slit experiment a laser produces intense coherent light as the light passes through the double slits it is diffracted spreading out into the space beyond this creates two overlapping sets of waves and an interference pattern is observed on a screen placed about a meter away we will focus on the interference pattern with within this specific region within this region the bright fringes exhibit equal intensity and consistent Fringe separation as shown the double slit experiment can be used to determine the wavelength Lambda of light the following three quantities have to be measured slit separation a in meters Fringe separation X in meters this is the distance between the centers of adjacent bright or dark fringes s flip to screen distance D in meters the formula for calculating the wavelength is Lambda equals ax / D this formula can explain why the Fringe separation for red light is greater than that for green and blue light as shown in the diagram given a constant slit separation a and slit to screen distance D the wavelength Lambda is directly proportional to The Fringe separation X since red light has the longest wavelength The Fringe separation X for red light is greatest blue light with the shortest wavelength exhibits the smallest Fringe separation exam style question five a double slit interference experiment is set up as shown fringes are formed on the screen the distance between successive bright fringes is found to be 4 m M given the wavelength of red light equals Lambda the slit separation equals a the slit to screen distance equals D and Fringe separation x = 4 mm as shown in the diagram two changes are then made to the experimental Arrangement the double slit is replaced by another double slit which has half the spacing the screen is moved so that its distance from the double slit is twice as great what is now the distance between successive bright fringes from the equation Lambda equal ax / D rearranging the equation as xal Lambda D / a from the diagram the old Fringe separation x = 4 mm substituting new D = 2D and new a = a / 2 so mux = 4 Lambda D / a substitute Lambda d/ a = 4 mm so newx = 16 mm exam style question six the diagram shows an arrangement for demonstrating two Source interference using coherent light of a single wavelength Lambda an interference pattern is observed on a screen 3.0 M away from the slits X and Y which have a separation of 1.0 mm the central bright Fringe is at Q and the second bright Fringe from the center is at P what is the distance between q and P the distance PQ equals 2x from the equation Lambda equal ax / D rearranging the equation as xal Lambda D / a substituting D equal 3 m and a = 1.0 * 10^ -3 M we get The Fringe separation x = 3 * 10^ 3 Lambda so 2x = 6 * 10^ 3 [Music] Lambda hey everyone welcome to PL Academy hope you found this video useful if you did it would be awesome if you could subscribe like the video share it with your friends and maybe drop a comment your support really helps me make more videos in this video I covered everything on the syllabus like you can see in the figure defraction grating a defraction grating consists of a large number of equally spaced lines ruled onto a glass or plastic slide each line acts as a source of defraction for the incident light the gratings can have up to 10,000 lines per cimer when light passes through the grating an interference pattern of fringes is observed in the diagram monochromatic light from a laser is incident normally on a transmission defraction grating beyond the grating interference fringes are formed and can be observed on a screen Sim similar to the double slit experiment the fringes are also referred to as Maxima the central Fringe is called the zeroth order maximum a0 the next Fringe is the first order maximum A1 the next Fringe is the second order maximum A2 and so on the pattern is symmetrical so there are two first order Maxima A1 2 second order Maxima A2 and so on as shown to analyze the pattern we measure the angles Theta and beta at which the Maxima are formed rather than measuring their separation with double slits the fringes are equally spaced and the angles are very small with a defraction grating the angles are much greater and the fringes are not equally spaced the differences in the interference patterns between defraction gratings and double slits are as follows brighter and sharper fringes defraction gratings typically produce brighter and sharper fringes than double slits because more light passes through the numerous slits in the grating larger Fringe separation defraction gratings generally exhibit larger Fringe separations than double slits because the slit spacing in gratings is much smaller than the slit separation in typical double slit experiments the wavelength Lambda can be determined with a defraction grating by measuring the angles at which the Maxima occur we can determine the wavelength of the incident light and calculating using the equation D sin theta equals n Lambda where D is the grating spacing in meters Theta is the angle between the zero order maximum and N order maximum n is the order of Maximum as 1 2 3 and so on and Lambda is the wavelength of light in meters if the grating has K line per meter then the grating spacing D can be calculated using d = 1 / K exam style question one a beam of red laser light of wavelength 633 nanom is incident normally on a defraction grating with 600 lines per millimet the beam of red light is now replaced by a beam of blue laser light of wavelength 445 nanom a replacement defraction grating is used so that the first order maximum of the blue light appears at the same position on the screen as the first order maximum of the red light from the original laser this means that the angle Theta is the same for both and the first order corresponds n equal 1 how many lines per millimeter are there in the replacement defraction grating using the equation D sin Theta equal n Lambda for the blue light and the replacement grating we have D2 sin Theta = 1 time blue Lambda for the red light and the original grating we have D1 sin Theta = 1 * red Lambda dividing the first equation by the second equation we get D2 sin Theta over D1 sin Theta equal blue Lambda over red Lambda so sin Theta is the same for both it cancels out leaving D2 / D1 equal blue Lambda over red Lambda substituting blue Lambda 445 nanom and red Lambda equal 633 nanom the nanometers cancel out so we get D2 = 445 over 633 * D1 substituting D2 = 1 / K2 and D1 = 1 / K1 where K2 and K1 are the number of line per millimeters for the replacement and original gratings respectively substituting K1 equals 600 per MIM we solve K2 = 850 per millim for two significant figures exam style question question two monochromatic light of wavelength 690 nanom passes through a defraction grating with 300 lines per millimet producing a series of Maxima bright spots on a screen what is the greatest number of Maxima that can be observed the defraction grating with 300 lines per millimet so grating spacing D equals 1 over 300 mm or 1 over 300 * 10 power -3 M using the equation D sin Theta equal n Lambda substituting d = 1 300 power -3 M Theta = 90° and Lambda = 690 * 10^ -9 M we solve for n order Maxima equal 4.83 but n must be a whole numbers so the maximum order is four so the greatest number number of Maxima that can be observed is 4 + 1 + 4 is equal to 9 Maxima which includes the zeroth order maximum and four Maxima on left and right hand sides as shown we substitute the angle theta equals 90° because this represents the maximum possible angle from the zero order maximum to the highest order maximum I hope you found this video helpful if you did I would be grateful if you would subscribe share like and leave a positive comment your support will encourage me to create more content thank you